China Mathematical Olympiad

Proof Through point $L$ draw perpendicular lines to $AB$ and to $AC$, the feet are $L^{\prime }$ and $L^{\prime \prime }$ respectively. Let $\angle LAB={\alpha }_1$, $\angle LAC={\alpha }_2$, $\angle L{F}_{2}A={\alpha }_3$, and $\angle L{E}_{1}A={\alpha }_4$. We have $$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}=\frac{L{L}^{\prime }}{L{L}^{\prime \prime }}=\frac{L{F}_2\sin {\alpha }_3}{L{E}_1\sin {\alpha }_4}.①$$ Draw line segments $D_1{F}_2$ and $D_2{E}_1$ (see Figure 1). Since $△L{D}_1{F}_2\sim △L{D}_2{E}_1$ we get

Figure 1 Figure 2 $$\frac{L{F}_2}{L{E}_1}=\frac{D_1{F}_2}{D_2{E}_1}.\stackrel{◯}{2}$$ Draw line segments $D_2{F}_1$ and $D_1{E}_2$ (see Figure 2). By using the sine rule we obtain $$\frac{\sin {\alpha }_3}{\sin {\alpha }_4}=\frac{D_2{F}_1}{D_1{E}_2}.\stackrel{◯}{3}$$ Substituting ② and ③ into ①, we have $$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}=\frac{D_1{F}_2}{D_2{E}_1}\cdot \frac{D_2{F}_1}{D_1{E}_2}.\stackrel{◯}{4}$$ Similarly, write $\angle BMC={\beta }_1$, $\angle MBA={\beta }_2$, $\angle NCA={\gamma }_1$, $\angle NCB={\gamma }_2$, and we get $$\frac{\sin {\beta }_1}{\sin {\beta }_2}=\frac{E_1{D}_2}{E_2{F}_1}\cdot \frac{E_2{D}_1}{E_1{F}_2},\stackrel{◯}{5}$$ $$\frac{\sin {\gamma }_1}{\sin {\gamma }_2}=\frac{F_1{E}_2}{F_2{D}_1}\cdot \frac{F_2{E}_1}{F_1{D}_2}.\stackrel{◯}{6}$$ Multiplying ④, ⑤ and ⑥, we obtain $$\frac{\sin {\alpha }_1}{\sin {\alpha }_2}\cdot \frac{\sin {\beta }_1}{\sin {\beta }_2}\cdot \frac{\sin {\gamma }_1}{\sin {\gamma }_2}=1.$$ Finally, according to the inverse of Ceva's Theorem, we know $AL$, $BM$ and $CN$ have a common point.